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Using the digits 2, 3, 4, 7 and 8, Carlos will form five-digit positive integers. Only the digit 2 can be used more than once in any of Carlos’ five-digit integers. How many distinct fivedigit positive integers are possible?

 May 29, 2020
 #1
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Cases:

 

No repetition of 2:

\(cases = {5!}\)

 

2 is present two times:

\(cases = {{5!} \over 2!}\)

 

2 is present three time:

\(cases = {{5!} \over 3!}\)

 

2 is present four time:

\(cases = {{5!} \over 4!}\)

 

2 is present five time:

\(cases = {{5!} \over 5!}\)

 

all cases:

\(total = {5!+{{5!} \over 2!}+{{5!} \over 3!}+{{5!} \over 4!}+{{5!} \over 5!}}\)

\(total = {206}\)

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 May 29, 2020
edited by Hephaestus  May 29, 2020
 #2
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A computer code calculates a total of 501 distinct permutations and begins with:

 

{{2, 2, 2, 2, 2}, {2, 2, 2, 2, 3}, {2, 2, 2, 2, 4}, {2, 2, 2, 2, 7}, {2, 2, 2, 2, 8}, {2, 2, 2, 3, 2}, {2, 2, 2, 3, 4}, {2, 2, 2, 3, 7}, {2, 2, 2, 3, 8}, {2, 2, 2, 4, 2}, {2, 2, 2, 4, 3}, {2, 2, 2, 4, 7}, {2, 2, 2, 4, 8}, {2, 2, 2, 7, 2}, ...........and ends with:

 {8, 4, 7, 3, 2}, {8, 7, 2, 2, 2}, {8, 7, 2, 2, 3}, {8, 7, 2, 2, 4}, {8, 7, 2, 3, 2}, {8, 7, 2, 3, 4}, {8, 7, 2, 4, 2}, {8, 7, 2, 4, 3}, {8, 7, 3, 2, 2}, {8, 7, 3, 2, 4}, {8, 7, 3, 4, 2}, {8, 7, 4, 2, 2}, {8, 7, 4, 2, 3}, {8, 7, 4, 3, 2} = 501 permutations.

 May 29, 2020
 #3
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OK, young person! If you don't believe it, here are the 16 distinct COMBINATIONS, which you must permute as follows:

 

[(2, 2, 2, 2, 2), (2, 2, 2, 2, 3), (2, 2, 2, 2, 4), (2, 2, 2, 2, 7), (2, 2, 2, 2, 8), (2, 2, 2, 3, 4), (2, 2, 2, 3, 7), (2, 2, 2, 3, 8), (2, 2, 2, 4, 7), (2, 2, 2, 4, 8), (2, 2, 2, 7, 8), (2, 2, 3, 4, 7), (2, 2, 3, 4, 8), (2, 2, 3, 7, 8), (2, 2, 4, 7, 8), (2, 3, 4, 7, 8)] >>Total = 16 distinct combinations.

[1 + 5*4 + 6*20 + 4*60 + 1*120] =501 distinct permutations.

 May 29, 2020

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